# Solving Problems Involving Systems of Equations

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• 0:05 Systems of Equations
• 0:50 Counting Practice
• 2:35 Money Practice
• 4:58 Percent Practice
• 7:42 Lesson Summary
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Lesson Transcript
Instructor: Jeff Calareso

Jeff teaches high school English, math and other subjects. He has a master's degree in writing and literature.

Did you ever encounter a problem where you seem to have lots of information, but still have a couple of critical numbers missing? In this lesson, we'll practice solving word problems that require us to set up systems of equations.

## Systems of Equations

Lots of things have systems. For example, there's a knitter's system of organizing yarn by color or a chef's system of chopping and organizing ingredients before cooking. Even when society breaks down, systems still matter. Let's say it's the zombie apocalypse. You'll stand a better chance of surviving if you use systems.

This includes not only systems of avoiding zombies. It also means systems of equations. A system of equations is a group of two or more equations with the same variables. There are many different kinds of word problems involving systems of equations. In this lesson, we'll focus on a few of the most common types of these problems. Remember, though, that the principles at work are the same in all of them.

## Counting Practice

Let's start with one that could save your life. It's the zombie apocalypse. You try to take shelter at Farmer Zed's place, but his barn is full of zombie animals. You know he had 20 animals, a mix of chickens and pigs. Your recon scout tells you he counts 50 legs, but can't see what they are. You hope there are more chickens, because zombie chickens are easier to kill. Zombie pigs? Not so much.

Okay, math can help. First, we need our variables. What don't we know? The number of chickens and pigs. Let's use c for chickens and p for pigs.

Next, we need our equations. We know c + p = 20. That's all the chickens and pigs combined. We know the chickens should have 2 legs and the pigs should have 4, supposing none were gnawed off by zombie ducks. So, 2c + 4p = 50.

To solve, let's rearrange this first equation to c = 20 - p, then substitute 20 - p for c in the second problem. This is the substitution method. So, that's 2(20 - p) + 4p = 50. We get 40 - 2p + 4p = 50. That becomes 2p = 10. So, p = 5.

Let's plug in 5 for p in c + p = 20. So, c = 15. Let's check our math by plugging p = 5 and c = 15 into the second equation. That's 2(15) + 4(5) = 50. 30 + 20 = 50. Okay, so 5 zombie pigs and 15 zombie chickens.

## Money Practice

Let's try a problem involving money. In this post-apocalyptic hellscape you call home, you need two things most of all: bullets and cookies. Look, it may be the end times, but you still have a sweet tooth. Fortunately, you encounter some delusional retail workers who are convinced the monetary system should still function. I guess the zombies aren't much different than Black Friday shoppers to them.

Unfortunately, their cash registers no longer work, so you have no receipt. But, you know you spent \$615.00 on a total of 135 \$3.00 bullets and \$8.00 cookies. How many of each did you get?

Let's make our variables b for bullets and c for cookies. That's good enough for me. What are our equations? You got 135 total items. So, b + c = 135. And, with a money problem, we can multiply the cost of each item times the number of the item to get its total cost. So, 3b is the cost of \$3.00 bullets, and 8c is the cost of \$8.00 cookies. You spent \$615.00, so 3b + 8c = 615.

Let's use substitution again. We can make b + c = 135 into b = 135 - c. Then, substitute into the equation to get 3(135 - c) + 8c = 615. That simplifies to 405 - 3c + 8c = 615. 8c - 3c is 5c. And, 615 - 405 is 210. So, 5c = 210. Divide by 5, and get c = 42.

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